What is the integral of (x^2+2)/(x(x^2-1)) ?

The integral of (x^2+2)/(x(x^2-1)) is 1/2 (-4ln(x)+3ln|x^2-1|)+C

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integral of (x^2+2)/(x(x^2-1))

\int \frac{x ^{2} + 2}{x ( x ^{2} - 1 )} d x

\frac{1}{2} (- 4 ln (x) + 3 ln x^{2} - 1) + C

Solution steps

\int \frac{x ^{2} + 2}{x ( x ^{2} - 1 )} d x

Apply u-substitution

= \int \frac{u + 2}{2 u ( u - 1 )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{u + 2}{u ( u - 1 )} d u

Take the partial fraction of

\frac{u + 2}{u ( u - 1 )} : - \frac{2}{u} + \frac{3}{u - 1}

= \frac{1}{2} \cdot \int - \frac{2}{u} + \frac{3}{u - 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (- \int \frac{2}{u} d u + \int \frac{3}{u - 1} d u)

\int \frac{2}{u} d u = 2 ln ∣ u ∣

\int \frac{3}{u - 1} d u = 3 ln ∣ u - 1 ∣

= \frac{1}{2} (- 2 ln ∣ u ∣ + 3 ln ∣ u - 1 ∣)

Substitute back

u = x^{2}

= \frac{1}{2} (- 2 ln x^{2} + 3 ln x^{2} - 1)

2 ln x^{2} = 4 ln (x)

= \frac{1}{2} (- 4 ln (x) + 3 ln x^{2} - 1)

Add a constant to the solution

= \frac{1}{2} (- 4 ln (x) + 3 ln x^{2} - 1) + C

What is the integral of (x^2+2)/(x(x^2-1)) ?
The integral of (x^2+2)/(x(x^2-1)) is 1/2 (-4ln(x)+3ln|x^2-1|)+C